Question

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Answer 1

Answer:

Let the number of red balls be "r" and

Number of white balls be "w".

given that , half the number of white balls is equal to one third of the number of red balls.

Hence ,

$\frac{w}{2} = \frac{r}{3} \\ \\ after \: cross \: multiplication \: we \: get \\ \\ 3w \: = 2r \\ \\ hence \\ \\ w = \frac{2r}{3}$

And also given that, Thrice the total number of balls exceed seven times the number of white balls by 6 .

We know that, total number of balls = r + w

So, 3(r + w ) - 7(w) = 6

Substitute the value of "w" here,

$3( \frac{2r}{3} + \frac{r}{1} ) - 7( \frac{2r}{3} ) = 6 \\ \\ 3( \frac{2r + 3r}{3} ) - ( \frac{14r}{3} ) = 6 \\ \\ 3( \frac{5r}{3} ) - \frac{14r}{3} = 6 \\ \\ \frac{15r}{3} - \frac{14r}{3} = 6 \\ \\ \frac{15r - 14r}{3} = 6 \\ \\ \frac{r}{3} = 6 \\ \\ r \: = 18 \\ \\ we \: know \: that \: w \: = \frac{2r}{3} \\ \\ w = \frac{2(18)}{3} \\ \\ w = \frac{36}{3} \\ \\ w = 12$

Therefore, there 18 red balls and 12 white balls.